When the solution set has positive dimension, find the number of the components and the dimension. I.e., an irreducible decomposition of the variety defined by the system. Ideally we should also get the parametrization of each component.

Multiplicity information would be nice but not necessary.

Or more technically, I'd like to decompose a variety defined by a binomial system into (irreducible) toric varieties and enumerate the characters.

Binomials.m2, a Maclaulay2 package (http://thomas-kahle.de/bpd.html) seems to be the best. But when applied to large systems with 50 or more variables, Binomials.m2 simply does not terminate within any reasonable amount of time.

I am familiar with the algorithms behind, but before I write my own, I would like to know if there is anything even better out there. Or maybe I'm using Binomials.m2 incorrectly?

If you're using the binomialPrimaryDecomposition function in the Binomials package, then it is doing two things, which may be contributing to the computation time, but seem not be relevant to you. First, it computes all primary components, even when one or more coordinates is zero. Second, it is finding the defining ideals of the components, as opposed to just the monomial parametrization. Have you tried the partialCharacter command? It's not exactly what you want, but I think it gets you most of the way towards a parametrization of the primary components in the torus.
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Dustin CartwrightFeb 19 '13 at 2:19

I'm writing a symbolic computation answer but you may want to try numerical algebraic geometry software such as Bertini.
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Thomas KahleFeb 19 '13 at 7:10

1 Answer
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Let $I$ be the ideal generated by your binomial equations after clearing denominators.

As a general rule with binomial equations, disregard coefficients in the first run (see here for why and how: http://arxiv.org/abs/1107.4699 ). I would first solve the system where all coefficients are set to be one, which means the equations should be of the form monomial minus monomial = 0. Once you have solved this, you can deform the result to a solution of your original problem.

Since you want to disregard any components contained in coordinate hyperplanes ("only nonzero solutions"), you are actually looking for the saturation $I : (\prod_i x_i)^\infty$. If you have it, then you are basically done since the saturation is a lattice ideal. In characteristic zero it is unmixed and has trivial primary decomposition given by extending characters (Corollary 2.2 in "Binomial ideals" [Eisenbud,Sturmfels, Duke Journal 84, 1996].

To compute the saturation $I : (\prod_i x_i)^\infty$ the fastest software is 4ti2 ( http://www.4ti2.de/ ). There the command markov does what you want to do. You would have to specify the exponent vectors of your equations in a '.lat' file and then run markov. Note that you can also do this from Macaulay2 using the FourTiTwo package.

To iterate, compared to computing generators of the saturation the rest is trivial. For instance finding parameterizations of the components of the saturation is basically just a Smith normal computation as in Chapter 7 of of the book "Combinatorial Commutative Algebra" [Miller/Sturmfels, Springer GTM].

Thank you for your answer! I'm reading your paper right now. About the last paragraph, I know I can find the parameterization (and hence dimension) from Smith normal form easily. But how can I find the number of component? First since I only want nonzero solutions, I can turn each binomial equation into the form "Laurent monomial = constant". With this, let $A$ be the integer matrix made from the exponents. After turning it into Smith Normal form i.e. $PAQ$ is diagonal, is the absolute value of the product of the nonzero entries on the diagonal the number of components?
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ssquiddFeb 19 '13 at 17:00

Yes it is this product. The reason is also in the Eisenbud Sturmfels paper. In their terminology you are asking: How many saturations does the character defined by your Laurent lattice ideal have. The answer is: The order of the finite group $\text{sat}(L)/L$ and the Smith normal form computes a canonical embedding of L into $ZZ^d$ from which you extract it.
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Thomas KahleFeb 20 '13 at 8:29